Question

IF
y = e^x (sinx + cos x) Find dy
dx​

Answer 1

Answer:

2 $e^{x}$ cos x

Step-by-step explanation:

$y = e^{x}(sin x + cos x)$

Using the formula :-

$\frac{duv}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}$

Here u = $e^{x}$ and v=(sin x + cos x)

$e^{x} \frac{d(sin x + cos x)}{x} + (sin x + cos x)\frac{d(e^{x} )}{x}$

= $e^{x}$ (cos x -sin x) + (sin x + cos x) $e^{x}$

(Because of differentiation of cos x = - sin x and differentiation of sin x = cos x )

=  $e^{x}$cos x - $e^{x}$ sin x + $e^{x}$sin x + $e^{x}$ cos x

= 2$e^{x}$cos x